//相交链表
//不带环，相交返回相交节点；不想交返回NULL；
//思路二：双指针
struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB) {
	if (!headA || !headB)
		return NULL;
	struct ListNode* cur1 = headA;
	struct ListNode* cur2 = headB;
	while (cur1 != cur2){
		cur1 = cur1 == NULL ? headB : cur1->next;
		cur2 = cur2 == NULL ? headA : cur2->next;
	}
	return cur1;
}
